\subsection{Exercice 3}
Potential inside a sphere:\\\\
\begin{enumerate}
  \item Find the Green function  \begin{math}
                             G(\vec {r}\ , \vec{r}')
                             \end{math}\ for the inner of a sphere with radius
                             R by using Dirichlet conditions.\\
We guess the green function as:\\ \\
\begin{math}
G(\vec{r}\ , \vec{r}')\ =\ \frac{1}{\vert \ \vec{r}-\vec{r}' \vert}-
\frac{\frac{q}{q_f}}{\vert \ \vec{r}-\vec{r}_f'  \vert}
\end{math}\\
(Its almost the same function as in the lecture)\\
Now we have to prove, that this is the searched green function by proving the
conditoins:
\begin{align}
 & 1)\ \nabla_{\vec{r}'}\ G(\vec{r}\ ,\vec{r}' )=\ -4\pi
\delta(\vec{r}-\vec{r}')\\
 & 2)\ G(\vec{r},\ \vec{r}'\epsilon \partial B_R)=\ 0
\end{align}\\
The first condition is ok, to prove the second we deform the green function a
bit:
\begin{align}
 G(\vec{r}\ , \vec{r}')	& = \frac{1}{\vert \ \vec{r}-\vec{r}'\vert}-\frac{\frac{q}{q_f}}{\vert \ \vec{r}-\vec{r}_f' 
 \vert}\\ 				& = \frac{1}{\sqrt{r^2+r'^2-2rr'\cos{\alpha}}}+\frac{\frac{q}{q_f}}{\sqrt{r^2+r_f'^2-2rr_f'\cos{\alpha}}} \end{align}\\
For Dirichlet condition we know:\\
\begin{align}
(\vert \vec{r} \vert \rightarrow R) \,
\frac{1}{\sqrt{R^2+r'^2-2Rr'\cos{\alpha}}}+\frac{\frac{q}{q_f}}{\sqrt{R^2+r_f'^2-2Rr_f'\cos{\alpha}}}
=\ 0 
\end{align}\\
This is true for all \begin{math} \alpha \end{math} and we get:
                                         
\begin{align}
\frac {\frac{1}{R}}{\sqrt{1+\left( \frac {r'}{R}
\right)^2-2\frac{r'}{R}\cos{\alpha}}}+\frac{\frac{q}{q_f}r_f' }{\sqrt{\left(
\frac{R}{r_f'} \right)^2+1-2\frac{R}{r_f'}\cos{\alpha} } }=\ 0
\end{align}\\
We compare these two terms an get the following because it have to be zero:\\
\begin{math}
\frac{R}{r_f'}\ =\frac{r'}{R} \,\,\,\, \rightarrow \,\, r_f'\ = \frac
{R^2}{r'}\\
\frac{1}{R}= -\frac{q}{q_f r_f'} \,\,\, \rightarrow \,\, q_f= -q\frac{r'}{R}
\end{math}\\
Using this \begin {math} q_f \end{math} and \begin {math} r_f'\end{math} in
equation (3) we get: 
\begin{align}
G(\vec{r}\ , \vec{r}')	& = \frac{1}{\vert \ \vec{r}-\vec{r}'\vert}-\frac{\frac{q}{-q\frac{r'}{R}}}{\vert \ \vec{r}-\frac {R^2}{r'} \vec{e_{r'}} \vert}\\
& = \frac{1}{\vert \ \vec{r}-\vec{r}'\vert}-\frac{\frac{R}{r'}}{\vert \ \vec{r}-\frac {R^2}{r'} \vec{e_{r'}} \vert}\\ 
& = \frac{1}{\vert \ \vec{r}-\vec{r}'\vert}-\frac{1}{\vert \frac{r'}{R}\vec{r}-\frac{R}{r'}\vec{r}'\vert}
\end{align} \\
This term is the proof for the second condition \begin{math}
                                                (r' \rightarrow R)
                                                \end{math} (2) we had to
prove, so this is the searched green function!
\item Consider a point charge q at $\vec r_0=(x_0,0,0)$ with $ |\vec r_0|<R $
  and calculate the electrostatic potential $\Phi(\vec r)$ inside the sphere.\\
  \\
  {\bf Calculation:}\\
  \begin{align*}
  &\Phi(\vec r)=\int_G d^3r' \cdot G(\vec r,
  \vec{r'})\rho(\vec{r'})-\int_{\partial G}d^2r' \cdot \frac{1}{4\pi}\Phi(\vec
  {r'})\frac{\partial G(\vec r, \vec{r'})}{n'}\\
  &=\int_Gd^3r' \cdot G(\vec r,\vec{r'})q \delta(x'-x_0)\delta(y')\delta(z')-0\\
  &=\int_Gd^3r' \cdot (\frac{1}{|\vec r-\vec{r'}|}\frac{1}{|\frac{r'\vec
  r}{R}-\frac{R\vec{r'}}{r'}|}) q \delta(x'-x_0)\delta(y')\delta(z')-0\\
  &=q(\frac{1}{|\vec r-\vec{r'}|}- \frac{1}{|\frac{x_0\vec
  r}{R}-\frac{R\vec{r'}}{x_0}|})
  \end{align*}
  \item Calculate the surface charge density $\sigma(|\vec r|=R)$ on the
  surface of the metal.\\ \\
  {\bf Calculation:}\\
  We know the identity: $\vec E(\vec r)=4\pi \sigma(\vec r)\vec n(\vec r) $.
  Let $\vec R$ be a vector with $|\vec R|=R$. So lets look at the electric field
  first:\\
  \begin{align*}
  &\vec E(\vec R)=-\vec\nabla\Phi(\vec R)\\
  &=q(-\frac{\vec R-\vec r_0}{|\vec R-\vec r_o|^3}+\frac{(\frac{r_0}{R})^2\vec
  R-\vec r_0}{|\frac{r_0\vec R}{R}-\frac{R\vec r_0}{r_0}|^3})\\
  &=\frac{q}{R}\frac{r_0^2-R^2}{|\vec r_0-\vec R|^3}\cdot \vec e_{\vec r}\\
  &\Rightarrow \sigma(\vec R)=\frac{q}{4\pi R}\frac{r_0^2-R^2}{|\vec r_0-\vec
  R|^3}\cdot \vec e_{\vec r}
  \end{align*}
  \item Calculate the force between the point charge and the metal sphere.\\ \\
  {\bf calculation:}\\
  The force is given as:
  \begin{align*}
  &\vec F(\vec R)=q\vec E(\vec R)\\
  &=q\frac{q}{R}\frac{r_0^2-R^2}{|\vec r_0-\vec R|^3}\cdot \vec e_{\vec R}
  \end{align*}
\end{enumerate}


